Example 2: eA
Compute eA, where
 A = 2 -2 1 -1 3 -1 -2 -4 3
Solution:

eA = T eD T -1 ; Diagonalize A:

We first use the equation det (A - I ) = 0 to find the eigenvalues:

 det (A - I ) = 2 - -2 1 -1 3 - -1 -2 -4 3 -
-Add the 2nd row to the 1st row:
 = 1 - 1 - 0 -1 3 - -1 -2 -4 3 -
 = ( 1 - ) 1 1 0 -1 3 - -1 -2 -4 3 -
-Take ( 1 - ) from the 1st row as a coefficient
-Add the 1st row to the 2nd row and multiplied by 2 to the 3rd row
 = (1-) 1 1 0 0 4 - -1 0 -2 3 -
 = (1-) 1 1 0 0 2 - 2 - 0 -2 3 -
 = (1-)(2-) 1 1 0 0 1 1 0 -2 3 -
 -Add the 3rd row to the 2nd row
 -Take ( 2 - ) from the 2nd row as a coefficient
 -Add the 2nd row multiplied by 2 to the 3rd row
 = ( 1 - ) ( 2 - ) 1 1 0 0 1 1 0 0 5 -
 = ( 1 - ) ( 2 - ) ( 5 - ) 1 1 0 = 0 0 1 1 0 0 1
 -Take ( 5 - ) from the 3rd row as a coefficient
 -The determinant of the upper triangular matrix is = 1·1·1 = 1
=> 1 = 1, 2 = 2, 3 = 5 (eigenvalues)
=> We have
 D = 1 0 0 0 2 0 0 0 5

Eigenvectors:

Using the equation (A - I ) = 0, we find the eigenvectors of A:

1 = 1: 1 = t ( 0, 1, 2 ), t C, t 0
2 = 2: 2 = t ( -1, 1, 2 ), t C, t 0
3 = 6: 3 = t ( 1, -1, 1 ), t C, r 0

Choose i's as follows:
1 = ( 0, 1, 2 ), 2 = ( -1, 1, 2 ), 3 = ( 1, -1, 1 )
and form the matrix T:

 T = 0 -1 -1 1 1 1 2 2 -1
We have
 T -1 = 3 3 0 -3 -2 1 0 -2 1
so we now calculate
 TeDT -1 = 0 -1 -1 1 1 1 2 2 -1
 e1 0 0 0 e2 0 0 0 e5
 3 3 0 -3 -2 1 0 -2 1
 = 3e2 2e2 + 2e5 e2 + e5 3e - 3e2 3e + 2e2 + 2e5 -e2 + e5 6e - 6e2 6e - 4e2 - 2e5 2e2 + e5
 7,4 -94,0 47,0 -4,7 96,7 -47,0 -9,3 -103,4 54,4

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